The Chain Rule for Partial Derivatives

Two reasons for z to change

Let z = f(x, y), and let x and y both depend on the time t. Then z depends on t too, and as t changes, z changes for two reasons at once: x is moving and y is moving.

Over a short time $\Delta t$Δt, x changes by about $\left(\frac{dx}{dt}\right)\Delta t$(dx/dt)Δt and y by about $\left(\frac{dy}{dt}\right)\Delta t$(dy/dt)Δt. The change in x alone raises z by about $\frac{\partial z}{\partial x}$∂z/∂x times the change in x, and the change in y alone raises z by about $\frac{\partial z}{\partial y}$∂z/∂y times the change in y. For small changes the two rises add.

One term per variable

Adding the two contributions and dividing by $\Delta t$Δt gives the chain rule: $\frac{dz}{dt}=\frac{\partial z}{\partial x}\cdot \frac{dx}{dt}+\frac{\partial z}{\partial y}\cdot \frac{dy}{dt}$dz/dt = ∂z/∂x · dx/dt + ∂z/∂y · dy/dt. There is one term for each variable that moves, and each term is the partial derivative of z with respect to that variable times the rate at which that variable changes.

As a tree, t branches to x and to y, and both branch to z. Multiply along each branch, then add the branches.

Take z = xy with $x=t^2$x = t² and y = 3t. Then $\frac{\partial z}{\partial x}=y$∂z/∂x = y and $\frac{\partial z}{\partial y}=x$∂z/∂y = x, while $\frac{dx}{dt}=2t$dx/dt = 2t and $\frac{dy}{dt}=3$dy/dt = 3. So $\frac{dz}{dt}=y\times 2t+x\times 3$dz/dt = y × 2t + x × 3. At t = 1, x = 1 and y = 3, so $\frac{dz}{dt}=3\times 2+1\times 3=9$dz/dt = 3 × 2 + 1 × 3 = 9.

Substituting first agrees: $z=t^2\times 3t=3t^3$z = t² × 3t = 3t³, so $\frac{dz}{dt}=9t^2$dz/dt = 9t², which is 9 at t = 1. The chain rule is needed when the substitution is long, or when only the rates at one moment are known.

A walker on the path x = 2 cos t, y = 1.5 sin t, over the bowl $z=x^2+\frac{y^2}{2}$z = x² + y²/2. The gold arrow is the gradient $\left(\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}\right)=\left(2x,y\right)$(∂z/∂x, ∂z/∂y) = (2x, y), and the chalk arrow is the velocity $\left(\frac{dx}{dt},\frac{dy}{dt}\right)$(dx/dt, dy/dt). At t = 0.6 the walker is at about (1.65, 0.85): the x term is 3.30 × (−1.13) = −3.73, the y term is 0.85 × 1.24 = 1.05, and $\frac{dz}{dt}=-2.68$dz/dt = −2.68. The walker is going downhill.

Reading the instrument

Along this path $z=4{\cos }^{2}t+1.125{\sin }^{2}t$z = 4 cos² t + 1.125 sin² t, and differentiating that directly gives $\frac{dz}{dt}=-2.875\sin 2t$dz/dt = −2.875 sin 2t, which is −2.68 at t = 0.6. The chain rule gives the same rate without writing z in terms of t.

Where the path runs along a contour of the bowl, the two terms cancel and $\frac{dz}{dt}=0$dz/dt = 0: the walker moves without climbing. That happens where sin 2t = 0, at the ends of the ellipse’s axes.

With one variable

If z depends on x alone, there is no second term, and the rule becomes $\frac{dz}{dt}=\frac{dz}{dx}\cdot \frac{dx}{dt}$dz/dt = dz/dx · dx/dt, the ordinary chain rule. The partial derivative is then an ordinary derivative, because nothing else is being held constant.

For $z=x^3$z = x³ with x = 2t + 1, $\frac{dz}{dt}=3x^2\times 2$dz/dt = 3x² × 2. At t = 1, x = 3, so $\frac{dz}{dt}=27\times 2=54$dz/dt = 27 × 2 = 54.

Two parameters

If x and y each depend on two variables s and t, then z does too, and each of its partial derivatives gets one term through x and one through y: $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial s}$∂z/∂s = ∂z/∂x · ∂x/∂s + ∂z/∂y · ∂y/∂s, and the same with t in place of s.

Take $z=x^{2}y$z = x²y with x = s + t and y = s − t, at s = 2 and t = 1, where x = 3 and y = 1. There $\frac{\partial z}{\partial x}=2xy=6$∂z/∂x = 2xy = 6 and $\frac{\partial z}{\partial y}=x^2=9$∂z/∂y = x² = 9. Since $\frac{\partial x}{\partial s}=1$∂x/∂s = 1 and $\frac{\partial y}{\partial s}=1$∂y/∂s = 1, $\frac{\partial z}{\partial s}=6+9=15$∂z/∂s = 6 + 9 = 15. Since $\frac{\partial x}{\partial t}=1$∂x/∂t = 1 and $\frac{\partial y}{\partial t}=-1$∂y/∂t = −1, $\frac{\partial z}{\partial t}=6-9=-3$∂z/∂t = 6 − 9 = −3.

The usual mistakes

Leaving out a term. If y is changing as well as x, then $\frac{\partial z}{\partial x}\cdot \frac{dx}{dt}$∂z/∂x · dx/dt is only part of $\frac{dz}{dt}$dz/dt.

Crossing the pairs. $\frac{\partial z}{\partial x}$∂z/∂x goes with $\frac{dx}{dt}$dx/dt, the rate of its own variable, never with $\frac{dy}{dt}$dy/dt.

Adding the rates without the partial derivatives. Each rate is weighted by how strongly z depends on that variable at that point.

A kite

In the application below, the length of a kite’s line is $L=\sqrt{x^2+y^2}$L = √(x² + y²), and the kite moves downwind and upward at the same time. The rate at which the line must be let out has one term from each motion.